Seven chits are numbered 1 to 7. Three are dr | vedclass

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(D) Let the numbers on the three chits be $X_1, X_2, X_3$. Each $X_i \in \{1, 2, 3, 4, 5, 6, 7\}$.We want the probability that $\min(X_1, X_2, X_3) =

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(D) Let the numbers on the three chits be $X_1, X_2, X_3$. Each $X_i \in \{1, 2, 3, 4, 5, 6, 7\}$.We want the probability that $\min(X_1, X_2, X_3) = 5$.This implies that all $X_i \ge 5$ $AND$ at least one $X_i = 5$.$P(\min(X_1, X_2, X_3) = 5) = P(\min(X_1, X_2, X_3) \ge 5) - P(\min(X_1, X_2, X_3) \ge 6)$.Since the draws are with replacement,the total number of outcomes is $7^3 = 343$.For $\min(X_i) \ge 5$,each $X_i$ must be in $\{5, 6, 7\}$. There are $3$ choices for each draw,so $3^3 = 27$ outcomes.For $\min(X_i) \ge 6$,each $X_i$ must be in $\{6, 7\}$. There are $2$ choices for each draw,so $2^3 = 8$ outcomes.Therefore,the number of favorable outcomes is $27 - 8 = 19$.The probability is $\frac{19}{343}$.

Seven chits are numbered $1$ to $7$. Three are drawn one by one with replacement. The probability that the least number on any selected chit is $5$,is

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